ANSWER COMPLETELY CORRECT AND YOU GET 5 STARS Two long, parallel wires carry cur

Question

ANSWER COMPLETELY CORRECT AND YOU GET 5 STARS

Two long, parallel wires carry currents of I1 = 3.10 A and I2 = 4.50 A in the directions indicated in the figure below, where d = 15.0 cm. Take the positive x direction to be to the right. Find the magnitude and direction of the magnetic field at a point midway between the wires, magnitude Your response differs from the correct answer by more than 10%. Double check your calculations. mu T direction degree counterclockwise from the +x axis Find the magnitude and direction of the magnetic field at point P, located d = 15.0 cm above the wire carrying the 4.50-A current, magnitude mu T direction degree counterclockwise from the +x axis

Explanation / Answer

B = u0 I / (2 pi R) = 2 * 10E-7 * I / R since u0 = 4 * pi * 10E-7

Both vectors make an angle of tan theta = (d/2 / d) = 1/2  with the x-axis

theta = 26.6 deg

Bx = -(B1 + B2) * cos 26.6

B1 + B2 = ( 2 * 10E-7 / (5^1/2 d / 2)) * (3.1 + 4.5) = 9.06 * 10E-6

Bx = -8.10 * 10E-6

By =  (B1 - B2) * sin 26.6

B1 - B2 = ( 2 * 10E-7 / (5^1/2 d / 2)) * (3.1 - 4.5) = - 1.67 * 10E-6

By = -7.48 * 10E-7

Tan theta = .748 / 8.1 = .092    theta = 5.27 below x-axis

theta = 185.3 deg

B = (8.1^2 + .748^2)^1/2 * 10E-6 = 8.13 * 10E-6 = 8.13 uT

The same approach is needed for part (b)

B2x = B2   and is directed along the negative x-axis (and B2y = 0)

B1 is located at 2^1/2 d from wire 1

B1 is directed at an angle of theta above the negative axis

I get theta = 45 deg

So B1 x = B1 cos 45   (along the negative x-axis)

B1y = B1 sin 45     and R = 2^1/2 d

Hope this helps.

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